$\frac{{\sin {{81}^o} + \cos {{81}^o}}}{{\sin {{81}^o} - \cos {{81}^o}}}$= 

${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $

$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ =

જો $\theta $ એ લઘુકોણ છે અને $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, તો $\tan \theta  = . . .$

$\frac{{4\sin {9^o}\sin {{21}^o}\sin {{39}^o}\sin {{51}^o}\sin {{69}^o}\sin {{81}^o}}}{{\sin {{54}^o}}}$ = 

જો $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ અને $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,કે જ્યાં $0 \le \alpha ,\beta \le \frac{\pi }{4}$. તો $\tan 2\alpha $ મેળવો.